Multiple Choice
What is the freezing point of a solution consisting of 88.5 g of sucrose (C12H22O11) in 282 g of water? Assume the freezing point depression constant (Kf) for water is 1.86 °C/m.
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14. Solutions
Freezing Point Depression
Multiple Choice
Calculate the freezing point of an aqueous solution with 95.21 g of barium chloride (BaCl2) dissolved in 1750 mL of water. Assume complete dissociation and use the freezing point depression constant (Kf) for water as 1.86 °C/m.
A
-0.93 °C
B
-5.58 °C
C
-1.86 °C
D
-3.72 °C
Verified step by step guidance1
First, calculate the molar mass of barium chloride (BaCl2). The molar mass is the sum of the atomic masses of barium (Ba) and two chlorine (Cl) atoms. Use the periodic table to find these values: Ba = 137.33 g/mol, Cl = 35.45 g/mol. Therefore, the molar mass of BaCl2 is 137.33 + 2(35.45) g/mol.
Next, determine the number of moles of BaCl2 in the solution. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Substitute the given mass of BaCl2 (95.21 g) and the calculated molar mass from the previous step.
Calculate the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent. First, convert the volume of water (1750 mL) to mass, assuming the density of water is 1 g/mL, so 1750 mL of water is 1750 g or 1.75 kg. Then, use the formula: \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \).
Determine the van't Hoff factor (i) for BaCl2. Since BaCl2 dissociates completely into three ions (Ba\(^{2+}\) and 2 Cl\(^{-}\)), the van't Hoff factor is 3.
Finally, calculate the freezing point depression using the formula: \( \Delta T_f = i \cdot K_f \cdot m \). Substitute the values for the van't Hoff factor (i = 3), the freezing point depression constant (Kf = 1.86 °C/m), and the molality calculated in the previous step. The result will give you the change in freezing point, which you subtract from the normal freezing point of water (0 °C) to find the new freezing point.
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